2092. Find All People With Secret
Read the full problem statement on LeetCode.
Difficulty: hard Acceptance: 45% Topics: Depth-First Search, Breadth-First Search, Union Find, Graph, Sorting
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(nlogn)
# Space: O(n)
import collections
class Solution(object):
def findAllPeople(self, n, meetings, firstPerson):
"""
:type n: int
:type meetings: List[List[int]]
:type firstPerson: int
:rtype: List[int]
"""
meetings.sort(key=lambda x: x[2])
result = {0, firstPerson}
adj = collections.defaultdict(list)
for i, (x, y, _) in enumerate(meetings):
adj[x].append(y)
adj[y].append(x)
if i+1 != len(meetings) and meetings[i+1][2] == meetings[i][2]:
continue
q = [i for i in adj.iterkeys() if i in result]
while q:
new_q = []
for u in q:
for v in adj[u]:
if v in result:
continue
result.add(v)
new_q.append(v)
q = new_q
adj = collections.defaultdict(list)
return list(result)
# Time: O(nlogn)
# Space: O(n)
import collections
class Solution2(object):
def findAllPeople(self, n, meetings, firstPerson):
"""
:type n: int
:type meetings: List[List[int]]
:type firstPerson: int
:rtype: List[int]
"""
meetings.sort(key=lambda x: x[2])
result = {0, firstPerson}
adj = collections.defaultdict(list)
for i, (x, y, _) in enumerate(meetings):
adj[x].append(y)
adj[y].append(x)
if i+1 != len(meetings) and meetings[i+1][2] == meetings[i][2]:
continue
stk = [i for i in adj.iterkeys() if i in result]
while stk:
u = stk.pop()
for v in adj[u]:
if v in result:
continue
result.add(v)
stk.append(v)
adj = collections.defaultdict(list)
return list(result)
# Time: O(nlogn)
# Space: O(n)
class UnionFind(object): # Time: O(n * alpha(n)), Space: O(n)
def __init__(self, n):
self.set = range(n)
self.rank = [0]*n
def find_set(self, x):
stk = []
while self.set[x] != x: # path compression
stk.append(x)
x = self.set[x]
while stk:
self.set[stk.pop()] = x
return x
def union_set(self, x, y):
x, y = self.find_set(x), self.find_set(y)
if x == y:
return False
if self.rank[x] > self.rank[y]: # union by rank
x, y = y, x
self.set[x] = self.set[y]
if self.rank[x] == self.rank[y]:
self.rank[y] += 1
return True
def reset(self, x):
self.set[x] = x
self.rank[x] = 0
class Solution3(object):
def findAllPeople(self, n, meetings, firstPerson):
"""
:type n: int
:type meetings: List[List[int]]
:type firstPerson: int
:rtype: List[int]
"""
meetings.sort(key=lambda x: x[2])
uf = UnionFind(n)
uf.union_set(0, firstPerson)
group = set()
for i, (x, y, _) in enumerate(meetings):
group.add(x)
group.add(y)
uf.union_set(x, y)
if i+1 != len(meetings) and meetings[i+1][2] == meetings[i][2]:
continue
while group:
x = group.pop()
if uf.find_set(x) != uf.find_set(0):
uf.reset(x)
return [i for i in xrange(n) if uf.find_set(i) == uf.find_set(0)]
Solution from kamyu104/LeetCode-Solutions · MIT
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